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One per cent composition of an organic compound $A$ is, carbon : $85.71 \%$ and hydrogen $14.29 \%$. Its vapour density is 14 . Consider the following reaction sequence
Identify $\underline{C}$,
Options:
Identify $\underline{C}$,
- A
- B $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
- C $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
- D $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
Solution:
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Verified Answer
The correct answer is:
$\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CO}_2 \mathrm{H}$
$\begin{aligned}
& \mathrm{C}=85.71 \% \quad=\frac{85.71}{12}=7.14 ; \quad \frac{7.14}{7.14}=1 \\
& \mathrm{H}=14.29 \% \quad=\frac{14.29}{1}=14.29 ; \quad \frac{14.29}{7.14}=2
\end{aligned}$
$\therefore \quad$ Empirical formula $=\mathrm{CH}_2$
and, empirical formula weight $=12+2=14$
Again, molecular formula weight
$\begin{aligned}
& =2 \times \text { vapour density } \\
& =2 \times 14=28 \\
\therefore \quad n & =\frac{28}{14}=2
\end{aligned}$
$\therefore$ Molecular formula $=\left(\mathrm{CH}_2\right)_2=\mathrm{C}_2 \mathrm{H}_4$
& \mathrm{C}=85.71 \% \quad=\frac{85.71}{12}=7.14 ; \quad \frac{7.14}{7.14}=1 \\
& \mathrm{H}=14.29 \% \quad=\frac{14.29}{1}=14.29 ; \quad \frac{14.29}{7.14}=2
\end{aligned}$
$\therefore \quad$ Empirical formula $=\mathrm{CH}_2$
and, empirical formula weight $=12+2=14$
Again, molecular formula weight
$\begin{aligned}
& =2 \times \text { vapour density } \\
& =2 \times 14=28 \\
\therefore \quad n & =\frac{28}{14}=2
\end{aligned}$
$\therefore$ Molecular formula $=\left(\mathrm{CH}_2\right)_2=\mathrm{C}_2 \mathrm{H}_4$
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